Notes on Diophantine Geometry Felipe Voloch and students June 5, 2008 Rational Points on Curves of Genus Zero An algebraic set over a field K is (the solution set of) a system of equations: f1 = 0 . X : . fm = 0 with f1, . , fm ∈ K[x1, . , xn]. Let X be an irreducible algebraic set, i.e., (f1, . , fm) is a prime ideal. Next, let R = K[x1, . , xn]/(f1, . , fm) and let K(X) denote the field of fractions of R. Then the dimension of X is the transcendence degree of K(X) over K: Tr degK K(X) = dim X. X is a curve if dim X = 1, that is, if some xi is not in K, and all other xj are algebraic over K(xi). We say that X, Y , irreducible algebraic sets, are birationally isomorphic if their function fields K(X),K(Y ) are isomorphic as field extensions of K. Example 1. If n = 2 and m = 1, f(x1, x2) = 0, then this is a curve. Now, we define the genus g(X) of an irreducible curve X. Over a subfield of C, the genus can be obtained from the topology of X(C). An algebraic definition of genus is the dimension of the space of regular 1-forms on a smooth projective model. If X is a plane curve of degree d, then we have the following formula for the genus: 1 X g(X) = (d − 1)(d − 2) − δ , 2 p p singular 1 where, if the singularity is an ordinary singularity with m branches, then 1 δp = 2 m(m − 1). (So, δp depends on the singularity. There is an algorithm for computing it in general, but you won’t find it in these notes.) 2 Background. Singular points are, for example, points (x, y) ∈ C with ∂f ∂f f(x, y) = (x, y) = (x, y) = 0. ∂x ∂y Example 2. For smooth (projective) plane curves of degree d, the genus is 1 2 (d − 1)(d − 2). In particular g = 0 ⇐⇒ d = 1, 2 g = 1 ⇐⇒ d = 3 g > 1 ⇐⇒ d ≥ 4. Definition. An irreducible curve X is parametrizable (also, rational) over K if there exist ϕ1, . , ϕn ∈ K(t) (not all constant) such that for all j = 1, . m we have fj(ϕ1, . , ϕn) = 0. Theorem 1. A curve is parametrizable over K if and only if it has genus zero and a smooth point with coordinates in K. Proof. (⇒) If X is parametrizable then K(X) embeds in K(t). This is seen by considering the map α : K[x1, . , xn] → K(t) which sends xi 7→ ϕi(t). Then fi ∈ ker α, so that α induces a map R = K[x1, . , xn]/(f1, . , fm) → K(t). Thus, by Luroth’s theorem, K(X) is purely transcendental over K. X is therefore birationally isomorphic to a line, and so has genus zero and many smooth points in K. (⇐) A sketch of the proof in this direction: Use g(X) = 0 together with the smooth point P and the Riemann-Roch space L(P ) of functions whose only pole is a simple pole at P to get that dim L(P ) = 2. Then, we find a 1 non-constant f ∈ L(P ), where f : X → P has only one (simple) pole, so it −1 1 has degree 1, making it a birational isomorphism. Thus, f : P → X is a parametrization. Example 3. Consider the curve X : x2 + y2 = 1. Then X has a point at P = (1, 0). By the degree formula for the genus, g(X) = 0. We will give a parametrization of X working geometrically: Any line through P intersects 2 the curve in exactly one other point (unless it is tangent at P ). The equation of a line through (1, 0) is given by y = t(x − 1). Substituting this into X for t2−1 −2t y and solving for x gives: x = t2+1 . Then, y = t(x − 1) = t2+1 . This is our parametrization. Assume K is perfect (e.g. has characteristic zero). By the genus for- mula and the fact that the genus is non negative, a cubic has at most one singularity. If a curve has a singular point in a field bigger than K, then its conjugates are also singular points so the curve must have other singular points. Therefore, we may conclude that an absolutely irreducible cubic with a singular point automatically has its singular point with coordinates in the ground field. The method used in the last example to find a parametrization works in this instance, too. You take the pencil of lines through the singular point, and because it is singular, this will give you a parametrization, hence lots of points. That is we get the following: Corollary 2. A singular cubic always has lots of points over the ground field. Example 4. Consider X : y2 = x3 + x2. Then X has a singular point at the origin P = (0, 0). The equation of a line through (0, 0) is given by y = tx. Then, we get x = t2 − 1, and so y = t(t2 − 1). 2 2 Example 5. But, x +y = −1 has no solutions over R. So, a smooth curve of degree 2 and genus 0, may have no points over Q or even R. However, it does have points over C. Example 6. Another example is x2 + y2 = 0. But, this curve is reducible: X = {x = iy} S {x = −iy}. Theorem 3. Every curve of genus zero over K is birationally isomorphic to a conic. Proof. Ww will not give a proof, just remark that the proof uses the negative of the canonical divisor, which gives us a divisor of degree two on the curve. 3 A conic is an absolutely irreducible curve of degree 2. Given by f(x, y) = 0 where deg f = 2. If we are looking for rational solutions of an equation f(x, y) = 0 where deg f = d, we can also look at the homogenous polynomial of degree d in x, y, z given by zdf(x/z, y/z). We have to be careful with z = 0. A conic can be further simplified by diagonalization. So we can reduce the study of conics to equations of the form ax2 + by2 + cz2 = 0 where absolute irreducibility is equivalent to abc 6= 0. 2 2 2 If a, b, c ∈ Q and we want to find out if ax + by + cz = 0 has solutions, we can assume that a, b, c ∈ Z by clearing denominators. We can also assume that a, b, c have no common factors. If a = m2 · a0 then ax2 + by2 + cz2 = 0 has a solution if and only if a0x2 + by2 + cz2 = 0 has a solution (replace x by mx). In this way we can assume that a, b, c are square free. Suppose there exists a prime p such that p divides both a and b. Then a = a0 · p and b = b0 · p. If ax2 + by2 + cz2 = 0 has a solution then pa0x2 + pb0y2 + cz2 = 0 a0(px)2 + b0(py)2 + pcz2 = 0 has a solution. Proceeding in this way, I can eliminate common factors of any two of a, b, c. Legendre’s Theorem. Suppose that a, b, c ∈ Z are nonzero, square free, and pairwise coprime. Then the equation ax2 + by2 + cz2 = 0 has a solu- 3 tion in Z −{(0, 0, 0)} if and only if the following two conditions are satisfied. (i) a, b, c are not all of the same sign. (ii) For all odd primes p | abc there is a solution to ax2 + by2 + cz2 ≡ 0 mod p where all of x, y, z 6≡ 0 mod p Proof. (⇑) If p | abc assume that p | a. We are asuming that ∃ x0, y0, z0 6≡ 2 2 2 2 2 0 mod p with ax0 + by0 + cz0 ≡ 0 mod p which is equivalent to by0 + cz0 ≡ 0 mod p. Let !2 b z u2 = − = 0 c y0 4 As a polynomial, ax2 + by2 + cz2 ≡ by2 + cz2 mod p ≡ c(z2 − u2y2) mod p ≡ c(z − uy)(z + uy) mod p 0 So for every odd prime p | abc there are linear forms Lp and Lp in x, y, z with 2 2 2 0 ax + by + cz ≡ LpLp mod p Also, for p = 2, ax2 + by2 + cz2 ≡ (ax + by + cz)2 mod 2 0 So there is also L2 and L2 such that 2 2 2 0 ax + by + cz ≡ L2L2 mod 2 By the Chinese Remainder Theorem, there exist linear forms L and L0such that ax2 + by2 + cz2 ≡ LL0 mod abc Consider integers x0, y0, z0 that satisfy the inequalities p p p 0 ≤ x0 ≤ |bc|, 0 ≤ y0 ≤ |ac|, 0 ≤ z0 ≤ |ab| Since a, b, c are square free and coprime, we actually have p p p 0 ≤ x0 < |bc|, 0 ≤ y0 < |ac|, 0 ≤ z0 < |ab| How many such triples are there? The answer is greater than ! ! ! j k j k j k 1 + p|bc| 1 + p|ac| 1 + p|ab| > |abc|. 5 By the Pigeonhole Principle there exists distinct such triples (x0, y0, z0) 0 0 0 and (x0, y0, z0) with 0 0 0 L(x0, y0, z0) ≡ L(x0, y0, z0)mod abc which implies that 0 0 0 L(x0 − x0, y0 − y0, z0 − z0) ≡ 0 mod abc and consequently 0 2 0 2 0 2 a(x0 − x0) + b(y0 − y0) + c(z0 − z0) ≡ 0 mod abc 0 0 0 Let x = x0 − x0, y = y0 − y0 and z = z0 − z0. Then x < p|bc|, y < p|ac|, z < p|ab| and ax2 + by2 + cz2 ≡ 0 mod abc By condition (i) we may suppose without a loss of generality that a < 0, b < 0, and c > 0.